Find the minimum and maximum indices of a list given a condition

Filter the zipped list with its indixes and take the min and the max:

>>> list_A = [0,0,0,1.0,2.0,3.0,2.0,1.0,0,0,0]>>> filtered_lst = [(x,y) for x,y in enumerate(list_A) if y > 0]>>> max(filtered_lst)(7, 1.0)>>> min(filtered_lst)(3, 1.0)

If you just need the index, unpack the returned value:

>>> maX,_ =  max(filtered_lst)>>> maX7

An alternative would be to use next():

list_A = [0,0,0,1.0,2.0,3.0,2.0,1.0,0,0,0]print(next(idx for idx, item in enumerate(list_A) if item>0))print(next(len(list_A)-1-idx for idx, item in enumerate(list_A[::-1]) if item>0))

Output

37

Using next() to find the first item in the list > 0 is an elegant solution.

To find the last item in the list > 0 is trickier with this method. I use next() to iterate over and find the first item > 0 in the reversed list using list_A[::-1]. I then convert the index generated to the correct index by subtracting it from len(list)-1, using len(list)-1-idx .

You can use the np.where function to return indices of all elements > 0

In [116]: list_A = [0,0,0,1.0,2.0,3.0,2.0,1.0,0,0,0]In [117]: arr = np.array(list_A)In [118]: indx = np.where(arr > 0)[0]In [119]: mini = indx[0]In [120]: miniOut[120]: 3In [121]: maxi = indx[-1]In [122]: maxiOut[122]: 7